FP3 June 2010 Q6
6. \[\mathbf{M} = \begin{pmatrix} 1 & 0 & 3 \\ 0 & -2 & 1 \\ k & 0 & 1 \end{pmatrix},\ \text{where } k \text{ is a constant.}\]
Given that \(\begin{pmatrix} 6 \\ 1 \\ 6 \end{pmatrix}\) is an eigenvector of \(\mathbf{M}\),
(a) find the eigenvalue of \(\mathbf{M}\) corresponding to \(\begin{pmatrix} 6 \\ 1 \\ 6 \end{pmatrix}\), (2)
(b) show that \(k = 3\), (2)
(c) show that \(\mathbf{M}\) has exactly two eigenvalues. (4)
A transformation \(T: \mathbb{R}^3 \to \mathbb{R}^3\) is represented by \(\mathbf{M}\).
The transformation \(T\) maps the line \(l_1\), with cartesian equations \(\dfrac{x - 2}{1} = \dfrac{y}{-3} = \dfrac{z + 1}{4}\), onto the line \(l_2\).
(d) Taking \(k = 3\), find cartesian equations of \(l_2\). (5)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 1 & 0 & 3 \\ 0 & -2 & 1 \\ k & 0 & 1 \end{pmatrix}\begin{pmatrix} 6 \\ 1 \\ 6 \end{pmatrix} = \lambda\begin{pmatrix} 6 \\ 1 \\ 6 \end{pmatrix}\) \(\begin{pmatrix} 24 \\ 4 \\ 6k + 6 \end{pmatrix} = \begin{pmatrix} 6\lambda \\ \lambda \\ 6\lambda \end{pmatrix}\) | |
| Uses the first or second row to obtain \(\lambda = 4\) | M1A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Uses the third row and their \(\lambda = 4\) to obtain \(6k + 6 = 24 \Rightarrow k = 3\) * | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix} 1 - \lambda & 0 & 3 \\ 0 & -2 - \lambda & 1 \\ 3 & 0 & 1 - \lambda \end{vmatrix} = 0\) \(\Rightarrow (1 - \lambda)\left((-2 - \lambda)(1 - \lambda) - 0\right) - 0\left(0(1 - \lambda) - 3\right) + 3\left(0 - 3(-2 - \lambda)\right) = 0\) | M1 A1 |
| \(\Rightarrow (1 - \lambda)(-2 - \lambda)(1 - \lambda) + 9(2 + \lambda) = (2 + \lambda)\left(9 - (1 - \lambda)^2\right) = 0\) \(\left(\lambda^3 - 12\lambda - 16 = 0\right)\) \(\Rightarrow (\lambda + 2)(\lambda^2 - 2\lambda - 8) = 0\) \(\Rightarrow (\lambda + 2)(\lambda + 2)(\lambda - 4) = 0\) | M1 |
| \(\lambda = -2, 4\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Parametric form of \(l_1\): \(\left(t + 2, -3t, 4t - 1\right)\) | M1 |
| \(\begin{pmatrix} 1 & 0 & 3 \\ 0 & -2 & 1 \\ 3 & 0 & 1 \end{pmatrix}\begin{pmatrix} t + 2 \\ -3t \\ 4t - 1 \end{pmatrix} = \begin{pmatrix} 13t - 1 \\ 10t - 1 \\ 7t + 5 \end{pmatrix}\) | M1 A1 |
| Cartesian equations of \(l_2\): \(\dfrac{x + 1}{13} = \dfrac{y + 1}{10} = \dfrac{z - 5}{7}\) | ddM1A1 |
| (5) | |
| (13 marks) |